Question: Multiply the following complex numbers: $({-5+2i}) \cdot ({3-3i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({-5+2i}) \cdot ({3-3i}) = $ $ ({-5} \cdot {3}) + ({-5} \cdot {-3}i) + ({2}i \cdot {3}) + ({2}i \cdot {-3}i) $ Then simplify the terms: $ (-15) + (15i) + (6i) + (-6 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -15 + (15 + 6)i - 6i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -15 + (15 + 6)i - (-6) $ The result is simplified: $ (-15 + 6) + (21i) = -9+21i $